# Newton's laws - Skier’s acceleration

Problem Statement:

A skier of mass m goes down a slope inclined at an angle α = 30º with friction. The coefficient of friction between the skier and the slope is μ = 0.15. Calculate the skier’s acceleration.  Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

Solution:

First we will draw a picture of the physical situation described in the problem statement. Then we will identify the forces acting in the problem. The skier is on the Earth, so he is subject to gravity. As he is supported by the inclined plane, the reaction to this plane and the friction force must also be taken into account. The different forces as well as the Cartesian axes with their positive sign are represented in the following figure.

Notice that the x-axis is aligned with the inclined plane (the slope). The reason for this choice is that the skier’s acceleration vector is parallel to this plane and therefore it will be easier to calculate the projections of Newton’s second law vectors. The final result does not depend on the orientation of the axes, so we choose them so that it is easier to solve the problem. To calculate the skier’s acceleration we are going to apply Newton’s second law: With the forces that we have represented this gives: Note that in Newton’s second law we are calculating a sum of vectors; that is, we are calculating the sum of the forces. Therefore in the above expression the vectors signs are always positive regardless of the direction to which they point. Knowledge is free, but servers are not. Please consider supporting us by disabling your ad blocker on YouPhysics. Thanks!

Newton’s second law is a vector equation, so we will do the vectors projections onto the axes of the reference frame. You can see for instance a representation of the projections of the weight vector on the Cartesian axes in the following figure:  We number the equations so that it is easier to refer to them later.

Equation (2) gives us the value of the reaction: Remember that the friction force is given by: So after substituting it in equation (1) we get: We substitute P by mg and after simplifying the expression we can deduce the acceleration: Note that the skier’s acceleration does not depend on its mass. After substituting the numerical values given in the statement we get: In the problem we have used g = 10 m/s-2

Do not forget to include the units in the results of the problems.

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